Problem: Solve for $x$ : $x^2 + 3x - 18 = 0$
The coefficient on the $x$ term is $3$ and the constant term is $-18$ , so we need to find two numbers that add up to $3$ and multiply to $-18$ The two numbers $6$ and $-3$ satisfy both conditions: $ {6} + {-3} = {3} $ $ {6} \times {-3} = {-18} $ $(x + {6}) (x {-3}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 6) (x -3) = 0$ $x + 6 = 0$ or $x - 3 = 0$ Thus, $x = -6$ and $x = 3$ are the solutions.